# Conservation of angular momentum (electromagnetism)

## Homework Statement

An infinite wire of linear charge density $\lambda$ lies on the $z$ axis. An insulating cylindrical shell of radius $R$ is concentric with the wire and can rotate freely about the $z$ axis. The charge per unit area on the cylinder is $\sigma = -\lambda/2\pi R$ while the mass per unit area is $\rho$. A magnetic field $\mathbf{B} = B\hat{z}$ fills the region of the cylinder.

At $t=0$ the cylinder is at rest and the magnetic field is reduced such that $\mathbf{B}(t) = Be^{-t/\tau}\hat{z}$.

Use conservation of angular momentum to find the angular velocity $\omega$ of the shell for $t\to\infty$.

Maxwell.

## The Attempt at a Solution

Inside the region of the shell, the magnetic field is initially just $\mathbf{B} = B\hat{z}$. The electric field from a wire is $\mathbf{E} = \lambda \hat{s}/2\pi \epsilon_{0} s$. Therefore, the momentum density is $\mathbf{p} = \epsilon_{0}\mathbf{E}\times\mathbf{B} = \lambda B \hat{\phi}/2\pi s$. The angular momentum density is $\mathbf{l} = \mathbf{s}\times \mathbf{p} = \lambda \hat{z}B/2\pi$. The total angular momentum per length is therefore $\mathbf{L} = \int \mathbf{l} \,\text{d}V = \frac{1}{2}\lambda B R^2 \hat{z}$.

Now, at $t \to \infty$, the ambient magnetic field goes to zero, so the angular momentum remaining is the angular momentum due to the spinning cylinder plus the angular momentum due to the magnetic field it's making and the electric field from the wire. When the cylinder is rotating, it's basically a solenoid, so the standard Amperian loop of hright $h$ gives $B_{z}h = \mu_{0} I = \mu_{0} \frac{\text{d}Q}{\text{d} t} = \mu_{0} \frac{\text{d}Q}{\text{d} A}\frac{\text{d}A}{\text{d} t} = \mu_{0} \sigma \frac{\text{d}A}{\text{d} t}$. But then, $\text{d} A = h \text{d} s = hR\text{d}\theta = hR\omega \text{d} t\implies \frac{\text{d}A}{\text{d} t} = h R \omega$ so $\mathbf{B} = \mu_{0}\sigma R \omega \hat{z} = -\frac{\mu_{0}}{2\pi}\lambda \omega \hat{z}$. This is just $\mathbf{B} = B' \hat{z}$ for $B' = -\frac{\mu_{0}}{2\pi}\lambda \omega$, so the result for the angular momentum of this field with the electric field still holds, and $\mathbf{L}_{\text{field}} =\frac{1}{2}\lambda B' R^2 \hat{z}$. Finally, via conservation of momentum, $\mathbf{L}_{0} = \mathbf{L}_{\text{field}} + \mathbf{L}_{\text{spin}}\implies \frac{1}{2}\lambda B R^2 = \frac{1}{2}\lambda B' R^2 + I \omega$. Also, the moment of inertia per length is $I = mR^2 = \rho \frac{A}{h} R^2 = 2\pi \rho R^4$, so we have $\omega = 2\pi B\lambda/(8\pi^2 \rho - \mu_{0}\lambda^2)$.

The issue is that I get I much simpler result via Faraday's Law, and since it was a lot simpler I assumed that this is somehow wrong.

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mfb
Mentor
You used I for two different things now.

The formula for I (as moment of inertia) does not look right. A/h is the surface area, not the interior - it is 2 pi R, not something with R^2. And then your result is I/h where h is a length in z-direction, as you express everything in "per length in z-direction".

That is a very interesting type of problem.

• 1 person
TSny
Homework Helper
Gold Member
so we have $\omega = 2\pi B\lambda/(8\pi^2 \rho - \mu_{0}\lambda^2)$.

This is what I got also, using the conservation of angular momentum. [EDIT: Actually, I see now that I got a factor of R that you don't have: $\omega = 2\pi B\lambda/[R(8\pi^2 \rho - \mu_{0}\lambda^2)]$. I think the missing factor of R in your expression is due to the mistake that mfb has pointed to.

[EDIT 2: After checking it over for the sixth time, I am no longer getting the factor of R in the denominator. So, I again agree with your answer. The dimensions seem to check, too. Sorry for the confusion and I hope I have it right now.] (Nope, See correction in post below)

The issue is that I get I much simpler result via Faraday's Law, and since it was a lot simpler I assumed that this is somehow wrong.

In using Faraday's law, did you take into account the changing B field due to the increasing speed of rotation of the charged cylinder?

In any realistic setup, you would have ##\rho >> \mu_o \lambda## and ##\omega## will be very small [see correction in post below]. So you could safely neglect the magnetic field of the spinning cylinder in both approaches.

Last edited:
TSny
Homework Helper
Gold Member
OK, feel free to shoot me. But now I'm getting

$\omega = 2\pi B\lambda/(8\pi^2 \rho R - \mu_{0}\lambda^2)$. So, there's a factor of R in just the first term of the denominator and I believe the dimensions of the two terms in the denominator are now consistent. This is what you will get if you take into account mfb's remarks.

The correct comparison is that ##\rho R>> \mu_{0}\lambda^2## in any realistic setup. Then you can neglect the second term in the denominator which is equivalent to not worrying about the magnetic field produced by the rotating cylinder.

• 1 person
You used I for two different things now.

The formula for I (as moment of inertia) does not look right. A/h is the surface area, not the interior - it is 2 pi R, not something with R^2. And then your result is I/h where h is a length in z-direction, as you express everything in "per length in z-direction".

That is a very interesting type of problem.

Ah yes, thank you! The dimensions were incorrect.

OK, feel free to shoot me. But now I'm getting

$\omega = 2\pi B\lambda/(8\pi^2 \rho R - \mu_{0}\lambda^2)$. So, there's a factor of R in just the first term of the denominator and I believe the dimensions of the two terms in the denominator are now consistent. This is what you will get if you take into account mfb's remarks.

The correct comparison is that ##\rho R>> \mu_{0}\lambda^2## in any realistic setup. Then you can neglect the second term in the denominator which is equivalent to not worrying about the magnetic field produced by the rotating cylinder.

Applying the correction, we find the same result. Also, no, I didn't account for the changing magnetic field as the cylinder speeds up; doing this with Faraday yields the same answer, so it's probably right. Thanks again to both of you 